Template parameters for template function

[rush.william]

Hello All!

There is following code:

template < class T >
void f(T const & value)
{
};

Is there any way to force user to specify template parameters
explicitly? I want to do the following:
f<int>(1); // Normal
f(1); // This should give an error

Any ideas?...

WBR

[Gianni Mariani]

ru**********@gmail.com wrote:

Hello All!

There is following code:

template < class T >
void f(T const & value)
{
};

Is there any way to force user to specify template parameters
explicitly? I want to do the following:
f<int>(1); // Normal
f(1); // This should give an error

Any ideas?...

Try this:
template < class T >
struct f_helper
{
typedef T type;
};
template < class T >
void f(typename f_helper<T>::type const & value)
{
};

The compiler is unable to deduce the type T from f_helper<T>::type.

[Markus Moll]

Hi

ru**********@gmail.com wrote:

Is there any way to force user to specify template parameters
explicitly? I want to do the following:
f<int>(1); // Normal
f(1); // This should give an error

Any ideas?...

You would have to make T undeducible. (word sounds wrong?)
You could for example use an indirection:

template<typename T>
struct f_helper
{
const T& value;
f_helper(const T& value) : value(value) {}
operator const T& () const { return value; }
};

template<typename T>
void f(const f_helper<T>& h)
{
const T& value = h;
// ...
}

Then f(1) fails to compile, while f<int>(1) will still compile.

Markus

[Gleb Golubitsky]

Gianni Mariani:

ru**********@gmail.com wrote:

Hello All!

There is following code:

template < class T >
void f(T const & value)
{
};

Is there any way to force user to specify template parameters
explicitly? I want to do the following:
f<int>(1); // Normal
f(1); // This should give an error

Any ideas?...

Try this:
template < class T >
struct f_helper
{
typedef T type;
};
template < class T >
void f(typename f_helper<T>::type const & value)
{
};

The compiler is unable to deduce the type T from f_helper<T>::type.

Thanks, it works!...

[Pete Becker]

Markus Moll wrote:

>
You would have to make T undeducible. (word sounds wrong?)

The standard doesn't use the word "deducible," so there's no direct help
there. But it does use the word "deduced" as in "deduced context", and
it uses the word "non-deduced" as its opposite.

--

-- Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com)
Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." (www.petebecker.com/tr1book)